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CM 197 Mechanics of Materials Chap 13: Shear Forces and Bending Moments in Beams

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## CM 197 Mechanics of Materials Chap 13: Shear Forces and Bending Moments in Beams

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CM 197Mechanics of Materials Chap 13: Shear Forces and Bending Moments in Beams

Professor Joe Greene

CSU, CHICO

Reference: Statics and Strength of Materials, 2nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill, Westerville, OH (1997)

CM 197

Chap 13: Shear Forces and Bending Moments

- Topics
- Introduction
- Types of Beams
- Types of Loading
- Beam Reactions
- Shear Force and Bending Moment in Beams
- Shear Force and Bending Moment Diagrams
- Relationships Among Load, Shear, and Moment
- Sketching Shear and Moment Diagrams Using Relationships
- Shear and Moment Formulas

Introduction

Plane of symmetry

- Beams are members that carry transverse loads and are subjected to bending.
- Any structural member subjected to bending may be referred to as a beam.
- Shafts, girders, stringers, floor beams, joists, etc.
- Objective of chapter
- Determine the internal forces at various sections along a beam.
- Review Types of beams and supports.
- Calculation of beam sections.
- Calculation of internal forces (shear and moments) in the beam.
- Produce graphical representations of shear and bending moments.
- Assumptions
- Beams are straight and of uniform cross sections that possess a vertical plane of symmetry. Fig 13-1
- Horizontal positions
- Subjected to forces applied in the vertical plane of symmetry.

Types of Supports for Beams

R

Rx

Ry

Rx

M

Ry

- Supports
- Roller supports
- Roller or link support resists motion of the beam only along the direction perpendicular to the plane of the support (along axis)
- Reaction at roller support acts along the known direction. Fig 13-2
- Have one unknown Reaction force, R
- Hinge supports
- Resists motion of the beam at the support in any direction on the plane of loading.
- Have two unknown Reaction forces- Rx and Ry
- Fixed supports
- Beam is built into wall or column.
- End of beam is fixed and doesn’t move.
- Have two unknown Reaction forces, Rx and Ry, and one unknown Moment, M

Types of Beams

- Beams
- Simple Beam (Statically Determinate- 3 unknowns)
- Beam supported at its end with a hinge and a roller
- Overhanging Beam (Statically Determinate- 3 unknowns)
- Simple supported beam with an overhang from one or both ends
- Cantilever Beam (Statically Determinate- 3 unknowns)
- Beam that is fixed at one end and free at the other
- Propped cantilever beam (Statically Indeterminate >3 unknowns)
- Beam that is fixed at one end and simply supported at the other.
- Fixed Beam (Statically Indeterminate >3 unknowns)
- Beam with both ends are fixed to supports
- Continuous Beam (Statically Indeterminate >3 unknowns)
- Beam is supported on a hinge and two or more roller supports

Types of Loading

- Loading
- Concentrated Loads
- Load is applied at a specific point on the beam and is considered a discrete force acting at the point.
- Example, weight fastened to a beam by a cable that applies a concentrated load.
- Uniform Loads
- Load is distributes over a part or the entire length of the beam.
- Load is force per unit length of beam (lb/ft or N/m)
- Replaced with equivalent force = uniform load times length of beam
- Example, weight of beam.
- Linearly Varying Loads
- Load is distributed with a uniform variation of intensity.
- Occurs on a vertical or inclined wall due to liquid pressure.
- Load changes along the length of the beam.

Beam Reactions

- Beam Reactions
- Review of Chapter 3
- Procedure
- Write three independent equilibrium equations to solve for the three unknowns.
- Three equations could be combinations of force-component or moment equations, as long as they are independent of each other.
- This chapter assume no horizontal forces.
- Thus reaction force in horizontal direction is zero.
- Assume, sum of forces in x direction is 0, thus the horizontal reaction is 0.
- Example, 13-1
- Example, 13-2

Shear Force & Bending Moment in Beams

100 lb

200 lb

Rx

1 ft

3 ft

Ry

R

10 ft

- Shear Force and Bending Moment
- Developed in a beam to resist the external forces and to maintain equilibrium.
- Figure 13-7
- Beam is 10 ft long with roller support on the right and hinge support on left.
- Beam is subjected to two concentrated loads.
- Results
- For cross section at 3 feet from the left, the forces and moments need to be balanced.
- Balance the forces and moments at each section
- Normal forces: Rx = 0; Ry =150 pounds;
- Shear force: V = 150 – 100 = 50 lbs; Moment is acting clockwise and = 3*150 – 100*2 = 250lb-ft;
- Internal resisting moment moment has to be equal and opposite of moment and acting opposite direction. M = 250ft-lb counterclockwise.
- For results for cross section from the right, the forces and moments are calculated in the same manner.

Beam Sign Conventions

- Positive Shear
- Shear force at a section is positive if the external forces on the beam produce a shear effect that ends to cause the left side of the section to move up relative to the right side. Fig 13-8a
- Positive Moment
- Bending moment at a section is positive if external forces on the beam produce a bending effect that causes the beam to concave upward at the section. Fig 13-8b
- Positive internal shear force, V, at a given section of a beam viewed from both directions is shown in Fig 13-8c.
- Positive internal moment M at a given section of a beam viewed from both directions is shown in Fig 13-8d.
- Rule 1 for finding shear forces
- Internal shear force at any section of a beam is equal to the algebraic sum of the external forces on either segment separated by the section.
- From the left of beam: upward force as positive
- From the right of the beam: downward force as positive
- Rule 2 for finding bending moments
- Internal bending moment at any section of a beam is equal to the algebraic sum of the moments about a section due to external forces.
- From either side: moment due to upward force as positive
- Example, 13-1 and Example, 13-2

Shear Force and Moment Diagrams

- Shear force and bending moment diagrams depict the variation of shear force and bending moment along a beam.
- Help visualize the shear forces and bending moments along beam
- Construct diagram
- Ordinate (y-axis) values are from balance of forces and moments.
- Compute reaction forces and moment couples.
- Plot values of the shear just below the free body diagram on the y-axis with positive values above the baseline and negative values below the baseline.
- Plot values of moments just below the shear diagram on same x-axis scale with positive values above the baseline and negative values below the baseline.
- Important aspects of diagrams
- For concentrated loads
- The shear diagram has a horizontal line between the loads
- The moment diagram has a line with constant slope
- For distributed loads
- The shear diagram has a line with a constant slope.
- The moment diagram has a curved line between moment points.
- If the shear diagram crosses the horizontal line (at zero shear) then it has a maximum or minimum moment at that point.
- Example 13-5

Relationships among Load, Shear, and Moment

- Certain relationships exist among the loading diagram, shear diagram and moment diagram.
- Assume incremental element of a beam.
- Relationships between Load and Shear
- For a distributed load, w, applied to a beam,
- The slope of the shear diagram (rate of change of the shear force per unit length) at any section is equal to the load intensity at that section.
- The shear force at a section is equal to the shear force at the previous section plus the total load between the two sections.
- Shear force at the section immediately to the right of a concentrated load is equal to the shear force immediately to the left plus the load.
- Shear force has an abrupt change at the concentrated load, upward = +
- Relationship between shear and moment
- Moment at a section is equal to the moment at the previous section plus the area under the shear diagram between the two sections.

Using Relationships of Diagrams

- Loading Diagram or free body diagram
- Show all applied forces and reactions on beam plus dimensions
- Never replace a distributed load with an equivalent concentrated load
- Shear diagram procedure
- Draw shear diagram directly below the loading diagram.
- Horizontal line is drawn at proper location below loading diagram.
- Draw vertical lines from controlling sections for supports and loads.
- Start at left end and compute shear at controlling sections using Eqn 13-5 VB = VA + LoadB-A
- Note: Section with concentrated force is applied, shear force diagram has an abrupt change at concentrated load location.
- Plot points on shear diagram using forces at each load.
- Connect points
- Note: Sections with distributed loads have a sloping line between them.
- Note: Slope of line is the intensity of the load. Concentrated loads have slope = 0

Using Relationships of Diagrams

- Moment diagram procedure
- Draw moment diagram directly below the shear diagram.
- Horizontal line is drawn at proper location below loading diagram.
- Draw vertical lines from controlling sections for supports and loads
- Calculate areas from shear diagram for each section.
- Note: moments at free end or ends of a simple beam are always = 0.
- Start at the left end and compute the moments using Eqn 13-7, MB = MA + AreaB-A
- Plot points on moment diagram using previous moments and areas at each section
- Connect points with straight lines (for concentrated loads) or curves (for distributed loads)
- Note: Maximum or minimum of moment occurs at the location where the shear is = 0 or where the shear changes sign.
- Example, 13-6, 13-7, 13-8, 13-9

Shear and Moment Formulas

- From previous section, shear force and bending moment diagram can be plotted for some simple but typical loading conditions.
- Table 13-1
- Procedure
- Match the loading conditions to one or more loading examples in Table 13-1
- Get the loading solutions from the example and use it in problem for
- Reaction forces, Maximum shear, maximum moments, etc.
- Method of superposition
- Effect of each load is computed separately and the then combined effect is added algebraically.
- Example, 13-10, 13-11

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